If $A\cong B$, then $A\otimes C\cong B\otimes C$. I think this is kind of true, since $\square\otimes C$ is a functor, so it preserves the isomorphism.
But what if we consider the example, $\mathbb{Z}\otimes\mathbb{Z}_2$ is not isomorphic to $2\mathbb{Z}\otimes\mathbb{Z}_2$. Because the latter is trivial while the first is not.
We can also consider the exact sequences:
$$0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_2\to0,$$
where the first map is simply multiple by 2,
and
$$0\to2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_2\to0,$$
where the first map is the inclusion.
If we tensor by $\mathbb{Z}_2$ we will get
$$\mathbb{Z}_2\to\mathbb{Z}_2\to\mathbb{Z}_2\to0$$
and 
$$(0\to)X\to\mathbb{Z}_2\to\mathbb{Z}_2\to0$$
respectively.
Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $r\otimes m$ is 0 in $\mathbb{Z}\otimes\mathbb{Z}_2$, then apparently it is 0 in $2\mathbb{Z}\otimes\mathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?
 A: $2\mathbb{Z}\otimes\mathbb{Z}_2$ is not trivial. It is trivial inside of $\mathbb Z \otimes \mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be
$$ 2n \otimes 0 = 0 \text{ and } 2n \otimes 1 = n \otimes 2 = 0 $$
but $n \otimes 2$ is only an element of $2\mathbb{Z}\otimes\mathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $\mathbb Z \otimes \mathbb{Z}_2$ to make sense of it.
Secondly, for your exact sequences. Do keep in mind that the isomorphism $2\mathbb{Z} \to \mathbb{Z}$ is a different map than the inclusion $2\mathbb{Z} \to \mathbb{Z}$.
So if we take the exact sequence 
$$ 2\mathbb{Z} \to \mathbb{Z}\to\mathbb{Z}_2\to 0 $$
where the leftmost map is inclusion and tensor with $\mathbb{Z}_2$, we get the following commutative diagram
$\require{AMScd}$
\begin{CD}
    2\mathbb{Z} \otimes \mathbb{Z}_2 @>\text{inclusion}>> \mathbb{Z} \otimes \mathbb{Z}_2 @>>> \mathbb{Z}_2\otimes \mathbb{Z}_2 @>>> 0 \\
    @VVV @VVV @VVV @VVV \\
    \mathbb{Z}_2 @>\times 2>> \mathbb{Z}_2 @>>> \mathbb{Z}_2 @>>>0
\end{CD}
where the rows are exact and the vertical arrows are isomorphisms.
That map $\mathbb{Z}_2 \xrightarrow{\times 2} \mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $\to \mathbb{Z}_2$ are
$$ \begin{array}{ccc}
2\mathbb{Z} \otimes \mathbb{Z}_2 & \longrightarrow & \mathbb{Z}_2 \\
2n \otimes 0 &\longmapsto & 0 \\
2n \otimes 1 & \longmapsto & n \bmod 2
\end{array}
\qquad \text{and} \qquad
\begin{array}{ccc}
\mathbb{Z} \otimes \mathbb{Z}_2 & \longrightarrow & \mathbb{Z}_2 \\
n \otimes 0 &\longmapsto & 0 \\
n \otimes 1 & \longmapsto & n \bmod 2
\end{array}  $$
So the map on the bottom we can compute by looking at the composition $\mathbb{Z}_2 \to 2\mathbb{Z} \otimes \mathbb{Z}_2 \to \mathbb{Z} \otimes \mathbb{Z}_2 \to \mathbb{Z}_2$ and this composition takes $1 \mapsto 2 \otimes 1 \mapsto 2 \otimes 1 \mapsto 0$.
A: $2\mathbb{Z}$ and $\mathbb{Z}_2$ are not isomorphic as $\mathbb{Z}$-modules. Take a generator of each module and multiply it by 2. 
